Determining averages with grouped data
The mean
We already know how to find the mean from a frequency table. Finding the mean for grouped data is very similar.
The grouped frequency table shows the number of albums bought by some students in the past year.
| Number of albums | Frequency |
| 0 - 4 | 10 |
| 5 - 9 | 12 |
| 10 - 14 | 6 |
| 15 - 19 | 2 |
| Number of albums | 0 - 4 |
|---|---|
| Frequency | 10 |
| Number of albums | 5 - 9 |
|---|---|
| Frequency | 12 |
| Number of albums | 10 - 14 |
|---|---|
| Frequency | 6 |
| Number of albums | 15 - 19 |
|---|---|
| Frequency | 2 |
- We know that 10 students have bought either 0, 1, 2, 3 or 4 albums, but we do not know exactly how many each student bought.
- If we assumed that each student bought 4 albums, it is likely that our estimate of the mean would be too big.
- If we assumed that each student bought 0 albums, it is likely that our estimate would be too small.
- It therefore seems sensible to use the mid-point of the group and assume that each student bought 2 albums.
Finding the mid-points of the other groups, we get:
| Number of albums | Frequency | Mid-point \(x\) | Mid-point\(x\) Frequency |
| 0 - 4 | 10 | 2 | 20 |
| 5 - 9 | 12 | 7 | 84 |
| 10 - 14 | 6 | 12 | 72 |
| 15 - 19 | 2 | 17 | 34 |
| Total = 30 | Total = 210 |
| Number of albums | 0 - 4 |
|---|---|
| Frequency | 10 |
| Mid-point \(x\) | 2 |
| Mid-point\(x\) Frequency | 20 |
| Number of albums | 5 - 9 |
|---|---|
| Frequency | 12 |
| Mid-point \(x\) | 7 |
| Mid-point\(x\) Frequency | 84 |
| Number of albums | 10 - 14 |
|---|---|
| Frequency | 6 |
| Mid-point \(x\) | 12 |
| Mid-point\(x\) Frequency | 72 |
| Number of albums | 15 - 19 |
|---|---|
| Frequency | 2 |
| Mid-point \(x\) | 17 |
| Mid-point\(x\) Frequency | 34 |
| Number of albums | |
|---|---|
| Frequency | Total = 30 |
| Mid-point \(x\) | |
| Mid-point\(x\) Frequency | Total = 210 |
The mean is \(\frac{{20 + 84 + 72 + 34}}{{10 + 12 + 6 + 2}} = \frac{{210}}{{30}} = 7\)
Remember: This is only an estimate of the mean.
The median
The median is the middle value when the values are arranged in order of size.
As the data has been grouped, we cannot find an exact value for the median, but we can find the group which contains the median.
| Number of albums | Frequency (f) |
| 0 - 4 | 10 |
| 5 - 9 | 12 |
| 10 - 14 | 6 |
| 15 - 19 | 2 |
| Number of albums | 0 - 4 |
|---|---|
| Frequency (f) | 10 |
| Number of albums | 5 - 9 |
|---|---|
| Frequency (f) | 12 |
| Number of albums | 10 - 14 |
|---|---|
| Frequency (f) | 6 |
| Number of albums | 15 - 19 |
|---|---|
| Frequency (f) | 2 |
There are 30 students, so we are looking for the group which contains the \((30 + 1) \div 2 = 15\frac{1}{2}th\) value, which lies halfway betwen the 15th and the 16th value.
The cumulative frequency for the first row is 10 and for the second row is 22.
The second row passes the the 15th and 16th value so the median must be in there.
The median is therefore within the class interval 5-9.
(There is an explanation of Cumulative Frequency on a previous page in this section)
The mode
The mode is the most common value.
In a frequency table with grouped data the best we can do is find the group with the highest frequency.
| Number of albums | Frequency (f) |
| 0-4 | 10 |
| 5-9 | 12 |
| 10-14 | 6 |
| 15-19 | 2 |
| Number of albums |
| Frequency (f) |
| 0-4 |
| 10 |
| 5-9 |
| 12 |
| 10-14 |
| 6 |
| 15-19 |
| 2 |
The modal group is 5-9.
In grouped frequency tables each group is sometimes referred to as a class of a class interval.
